Hybridisation of Cr2O7^2-

Cr2O7^2- is a polyatomic ion consisting of two chromium atoms and seven oxygen atoms. The hybridisation of Cr2O7^2- can be determined by following the below steps:

Step 1: Count the number of valence electrons present in Cr2O7^2-

- Chromium (Cr) has an atomic number of 24 and has 4 valence electrons.
- Oxygen (O) has an atomic number of 8 and has 6 valence electrons.
- Since Cr2O7^2- has two chromium atoms and seven oxygen atoms, the total number of valence electrons is 2(4) + 7(6) + 2 = 46.

Step 2: Determine the central atom

- In Cr2O7^2-, the two chromium atoms are bound to each other and the seven oxygen atoms are bound to the chromium atoms.
- Since there are only two chromium atoms, either of them can be considered as the central atom. However, the one with the higher oxidation state is usually considered as the central atom.
- In Cr2O7^2-, the oxidation state of each chromium atom is +6. Hence, either of the chromium atoms can be considered as the central atom.

Step 3: Determine the hybridisation of the central atom

- The hybridisation of the central atom can be determined using the formula: Hybridisation = (number of sigma bonds + number of lone pairs on the central atom)
- In Cr2O7^2-, the central chromium atom is bound to three oxygen atoms through sigma bonds and has one lone pair of electrons.
- Hence, the hybridisation of the central chromium atom can be calculated as:

Hybridisation = (3 + 1) = 4

- Therefore, the hybridisation of the central chromium atom in Cr2O7^2- is sp3.

Conclusion: The hybridisation of the central chromium atom in Cr2O7^2- is sp3.